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Answer:
a) if the two trains are going to collide
b) x = 538 m
Explanation:
To solve this exercise we use the kinematics relations in one dimension.
We set a reference system at the starting point of the passenger train
Freight train, we use index 1 for this train
x₁ = x₀ + v₁ t
passenger train
x₂ = v₂ t - ½ a₂ t²
as we are using the same system to measure the position of the two trains, at the meeting point the position must be the same
x₁ = x₂
we substitute
x₀ + v₁ t = v₂ t - ½ a₂ t²
½ a₂ t² + t (v₁ -v₂) + x₀ = 0
we subjugate the values
½ 0.1 t² + t (15-25) + 200 = 0
0.05 t² - 10 t + 200 = 0
t² - 200 t + 4000 = 0
we solve the second degree system
t = [200 ±[tex]\sqrt{200^2 - 4 \ 4000}[/tex] ]/ 2
t = [200 ± 154.9] / 2
t₁ = 177.45 s
t₂ = 22.55 s
therefore for the smallest time the two trains must meet t₂ = 22.55 s
merchandise train
x = 200+ 15 22.55
x = 538 m
passenger train
x = 25 22.55 -1/2 0.100 22.55²
x = 538 m
we see that the trains meet for this distance
a) if the two trains are going to collide
b) x = 538 m
c) see attachment for schematic graphics
